Let $i = \sqrt{\mathbf{-1}}$ we can define the extension $\mathbb{Q}(i)/\mathbb{Q}$. Then we have that: $$ \sqrt{\mathbf{-1}} \in \mathbb{Q}(i) \otimes \mathbb{R}$$
but what is that ring on the RHS? We could go even further and define: $$ [\sqrt{\mathbf{-1}}]_\infty = \left[ \begin{array}{cc} 1 & 0 \\ 0 & \sqrt{\mathbf{-1}}\end{array}\right] \in \mathrm{PGL}_2(\mathbb{Q}(i) \otimes \mathbb{R}) $$ This group makes even less sense if I can't compute the tenor product ring $\mathbb{Q}\otimes \mathbb{R}$ (or it's generalization).
There's ambiguity if the tensor product is $\otimes_\mathbb{Z}$ or $\otimes_\mathbb{Q}$ or $\otimes_\mathbb{R}$, however I did find this relation:
$$ \mathbb{Q}(\sqrt{-1}) \otimes \mathbb{R} \subset \mathbb{A}_{\mathbb{Q}(\sqrt{-1})}$$
Then it might be tensor product over $\mathbb{Q}$ as the Wikipedia page on adèles has:
$$ \mathbb{A}_F = \mathbb{A}_\mathbb{Q} \otimes_\mathbb{Q} F \simeq \Big( (\hat{\mathbb{Z}} \times \mathbb{R} ) \otimes_\mathbb{Z} \mathbb{Q} \Big) \otimes_\mathbb{Q} F $$ and that object is rather comlicated since I don't know if $\otimes_\mathbb{Z}$ and $\otimes_\mathbb{Q}$ can be easily switched around.
(A) $\mathbb{Q}(i) \otimes_\mathbb Q \mathbb{R}=\frac {\mathbb{Q}[X]}{(X^2+1)}\otimes_\mathbb Q \mathbb{R}=\frac {\mathbb{R}[X]}{(X^2+1)}=\mathbb C$
(B) $\mathbb{Q}(i) \otimes_\mathbb Z \mathbb{R}\stackrel {\bigstar}{=}\mathbb{Q}(i) \otimes_\mathbb Q \mathbb{R} \stackrel {\operatorname {(A)}}{=}\mathbb C$
(i) The genuine reason for the equality $\bigstar$ is the calculation $$\frac {1} {17}q \otimes_\mathbb Z r=\frac {q} {17}\otimes_\mathbb Z 17\frac {r}{17}=17\frac {q} {17}\otimes_\mathbb Z \frac {r}{17}=q\otimes_\mathbb Z\frac {1}{17}r$$ which proves that we can move $\frac {1} {17}$ from the left of $\otimes_\mathbb Z$ to the right even though a priori only integers can be moved from the left to the right of $\otimes_\mathbb Z$.
[As is well known, $17$ denotes an arbitrary positive integer]
(ii) The formal reason for equality $\bigstar$ is ...that there is a proof !
For example in Bourbaki, Commutative Algebra Chapter II, §2.7, Proposition 18, page 75.